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20 days ago

7

The surface is tilted to an angle of 37 degrees from the horizontal, as shown above in Figure 3. The blocks are each given a pus

h so that at the instant shown, they are moving toward each other. ii. At the instant shown in Figure 3, the blocks are moving toward each other with the same speed of 0.35m/s . The blocks collide 0.50 seconds later. What is the speed of the two-block system’s center of mass just before the blocks collide?

1 answer:

inna [2.2K]20 days ago

6 0

Answer:

Incomplete question: "Each block has a mass of 0.2 kg"

The velocity of the center of mass for the two-block system just prior to their collision is 2.9489 m/s

Explanation:

Provided information:

θ = angle of the surface = 37°

m = mass of each block = 0.2 kg

v = speed = 0.35 m/s

t = collision time = 0.5 s

Question: What is the velocity of the center of mass for the two-block system right before the blocks collide, vf =?

Change in momentum:

$delta(P)=F*delta(t)$

$P_{f} -P_{i}=F*delta(t)$

$2m(v_{f} -v_{i})=F*delta(t)$

$v_{i} =0.35-0.35=0$

It’s essential to calculate the required force:

$F=(m+m)*g*sin\theta$

Here, g = acceleration due to gravity = 9.8 m/s²

$F=(0.2+0.2)*9.8*sin37=2.3591N$

$v_{f} =\frac{F*delta(t)}{2m} =\frac{2.3591*0.5}{2*0.2} =2.9489m/s$

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Answer:

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Explanation:

Provided details include:

Wind speed = 75.0 km/hr

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18 days ago

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Jeff's body contains about 5.46 L of blood that has a density of 1060 kg/m3. Approximately 45.0% (by mass) of the blood is cells

ValentinkaMS [2425]

Answer:

a) Blood mass is $m= 5.7876kg$

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Explanation:

From the problem statement, we learn that

The blood volume is $V_b = 5.46 \ L = \frac{5.46}{1000} = 0.00546m^3$

The density of blood is $\rho_b = 1060 kg/m^3$

% of blood which consists of cells is = 45.0%

the % of blood that is plasma is = 55.0%

density of blood cells is $\rho_d = 1125kg/m^3$

% of cells that are white is = 1%

% of cells that are red is = 99%

The red blood cell diameter is = $7.5 \mu m = 7.5*10^{-6}m$

The red blood cell radius is $= \frac{7.5*10^{-6}}{2} = 3.75*10^{-6}m$

The mass is generally represented mathematically as

$m = \rho_b * V_b$

Substituting values

$m = 1060 * 0.00546$

$m= 5.7876kg$

Cell mass is $m_c$ = 45% of m

= 0.45 * 5,7876

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The volume of cells is $V_c$ = $\frac{m_c}{\rho_d}$

$= \frac{2.60442}{1125}$

$= 2.315 *10^{-3} m^3$

The white blood cells volume is $V_w$ = 1% of the cells volume

$= \frac{1}{100} * 2.315*10^{-3}$

$= 2.315*10^{-5}m^3$

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$= 2.21*10^{-16}m^3$

The red blood cells volume is $V_r = V_c - V_w$

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$= 2.29*10^{-3}m^3$

The total red blood cell count is $= \frac{V_r}{V_s}$

$= \frac{2.29 *10^{-3}}{2.21*10^{-16}}$

$= 1.037*10^{13}$

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$= \frac{2.315 * 10^{-5}}{2.21*10^{-16}}$

$= 1.04*10^{11}$

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$N_t=1.04*10^{13}$

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Answer:

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Explanation:

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diameter = 100 mm = 0.1 m

thickness = 0.1 mm = 1 × $10^{-4}$ m

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temperature of air = 15°C

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time needed for the puck to reduce its speed by 10%

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assuming initial velocity = v

therefore final velocity = 0.9v

implying a change in velocity is du = v

and clearance dy = h

shear stress acting on the surface is expressed as

= µ $\frac{du}{dy}$

therefore

= µ $\frac{v}{h}$ ............1

substituting the values

= 1.75 × $10^{-5}$ × $\frac{v}{10^{-4}}$

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the area between the air and puck is given by

Area = $\frac{\pi }{4} d^{2}$

area = $\frac{\pi }{4} 0.1^{2}$

area = 7.85 × $\frac{v}{10^{-3}}$ m²

thus, the force on the puck can be represented as

Force = × area

force = 0.175 v × 7.85 × $10^{-3}$

force = 1.374 × $10^{-3}$ v

now applying Newton's second law

force = mass × acceleration

- force = $mass \frac{dv}{dt}$

- 1.374 × $10^{-3}$ v = $0.03 \frac{0.9v - v }{t}$

solving for t = $\frac{0.1 v * 0.03}{1.37*10^{-3} v}$

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